Simple summation

Algebra Level 3

Find the following sum for ∣ n ∣ > 1 |n|>1 ∑ k = 0 ∞ n − k = ? \large \sum_{k=0}^{\infty}n^{-k} =\, ?

n n − 1 \dfrac{n}{ n-1} n n + 1 \dfrac{n}{ n+1} 1 2 n \dfrac{1}{ 2n} n ! n!

Akshay Sreenivasan by Akshay Sreenivasan , 23, India

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1 solution

Let S= 1 n 0 \frac{1}{n^{0}} + 1 n 1 \frac{1}{n^{1}} + 1 n 2 \frac{1}{n^{2}} + 1 n 3 \frac{1}{n^{3}} +...

=>S= 1 1 \frac{1}{1} + 1 n \frac{1}{n} + 1 n 2 \frac{1}{n^{2}} + 1 n 3 \frac{1}{n^{3}} +...

=>S-1= 1 n \frac{1}{n} + 1 n 2 \frac{1}{n^{2}} + 1 n 3 \frac{1}{n^{3}} +...

Taking the common term 1 n \frac{1}{n} outside, we get

S-1= 1 n \frac{1}{n} x [1+ 1 n \frac{1}{n} + 1 n 2 \frac{1}{n^{2}} +...]

The term inside the box=S

S-1= 1 n \frac{1}{n} x S

S- S n \frac{S}{n} =1

Solving it we get S= n n − 1 \frac{n}{n-1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

This is right if n > 1, but what happen if 0 < n ≤ 1 0 < n \leq 1 ?

Guillermo Templado - 5 years, 3 months ago

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